Interquartile Range, Mean Deviation about Mean and fitting of of Negative Exponential Distribution and Laplace Distribution

Interquartile Range of Negative Exponential Distribution:

Let X ~ Expo (θ). Then the i^th quartile (Q_i) of X is given by

$$\int_0^{Q_i} \ f(x) \ dx \ = \ \frac{i}{4} \ \ \ \ \ \ \ \ \ for \ i \ = \ 1, \ 2, \ 3$$

∴ The first quartile Q₁ is given by

$$\int_0^{Q_1} \ f(x) \ = \ \frac{1}{4}$$

$$\Rightarrow \ \int_0^{Q_1} \ \theta \ {e^-}^{\theta x} \ dx \ = \ \frac{1}{4}$$

$$\Rightarrow \ \left [ {-e^-}^{\theta x} \ \right ]_0^{Q_1} \ = \ \frac{1}{4}$$

$$\Rightarrow \ 1 \ - \ {e^-}^{\theta Q_1} \ = \ \frac{1}{4}$$

$$\Rightarrow \ {e^-}^{\theta Q_1} \ = \ \frac{3}{4}$$

$$\Rightarrow \ - \ \theta \ Q_1 \ = \ log_e \ \frac{3}{4}$$

$$\therefore \ Q_1 \ = \ - \ \frac{1}{\theta} \ (log_e 3 \ - \ log_e 4 )$$

Similarly, the third quartile Q₃ is given by

$$\int_0^{Q_3} \ \theta \ {e^-}^{\theta x} \ dx \ = \ \frac{3}{4}$$

$$\Rightarrow \ - \ \theta Q_3 \ = \ log_e \ \frac{1}{4}$$

$$\therefore \ Q_3 \ = \ \frac{1}{\theta} \ log_e 4$$

Then, the interquartile range is given by

$$Q_3 \ - \ Q_1 \ = \ \frac{1}{\theta} \ log_e \ - \ \left [ - \ \frac{1}{\theta} \ (log_e 3 \ - \ log_e 4) \ \right ]$$

$$Q_3 \ - \ Q_1 \ = \ \frac{1}{\theta} \ log_e 3$$

Mean Deviation about Mean of Negative Exponential Distribution

Let X ~ Expo (θ). Then

$$E(X) \ = \ \frac{1}{\theta}$$

The mean deviation about mean is given by

M.D. (mean) = E ( |X - E(X) | )

$$= \ \int_0^{\infty} \left | \ X \ - \ \frac{1}{\theta} \ \right | \ \theta \ {e^-}^{\theta x} \ dx$$

$$= \ \int_0^{\infty} \ | \ \theta x \ - \ 1 \ | \ \theta \ {e^-}^{\theta x} \ dx$$

$$= \ \int_0^{\infty} \ | \ y \ - \ 1 \ | \ {e^-}^{y} \ \frac{dy}{\theta} \ \ \ \ \ \ where \ y \ = \ \theta x \ and \ dy \ = \ \theta \ dx$$

$$= \ \frac{1}{\theta} \ \left [ \ \int_0^{1} \ (1-y) \ {e^-}^{y} \ dy \ + \ \int_{0}^{1} \ (y-1) \ {e^-}^{y} \ dy \ \right ]$$

$$= \ \frac{1}{\theta} \ [ \ {e^-}^{1} \ + \ {e^-}^{1} \ ] \ = \ \frac{2}{\theta \ e}$$

Fitting of Negative Exponential Distribution

In order to fit the negative exponential distribution to the given data, the steps to be followed are given below:

Step 1 :

Find the mean from the given data or distribution.

Step 2 :

Estimate the parameterθ of the negative exponential distribution by equating the mean and1/θ, the mean of the negative exponential dstribution to be fitted.That is making

$$\overline{x} \ = \ \frac{1}{\theta} \ \Rightarrow \ \overline{\theta} \ = \ \frac{1}{\overline{x}}$$

Step 3 :
Compute the area under the negative exponential curve for each class intervals (a,b) i.e. compute the probability that the negative exponential random variables X lies in the interval (a,b) as

$$P(a \ < \ X \ < \ b) \ = \ \int_a^{b} \ \theta \ {e^-}^{\theta x} \ dx$$

where, a≥ 0 and b > 0.

Step 4 :

Calculate the expected negative exponential frequencies by

f_e = N P (a < X < B)

Where, N =∑ f = Total observed frequency of the given data.

Uses of Negative Exponential Distribution

1. Negative exponential distribution has a great application in queuing theory and reliability theory
2. It is most useful model to study the waiting time until the first change or the lifetime of a certain electronic equipment and other machines i.e. the time until a component of equipment fails, the time taken to complete a job, the time taken to get a customer, the time between arrivals at a certain service station etc.
3. As time taken to decay radioactive particles follows the negative exponential distribution, it is most useful distribution in studying the decay time of a radioactive particles.

Laplace Distribution

The double exponential distribution with the general probability density function (pdf)

$$f(x) \ = \ \frac{1}{2} \ \theta \ {e^-}^{\theta |x- \mu|} \ \ \ \ \ ; \ - \infty \ < \ x \ < \ \infty, \ - \infty \ < \ \mu \ < \ \infty, \ \theta \ > \ 0$$

is called Laplace distribution with parameterμ andθ.

Another general form of laplace distribution is defined as follows:

A continious random variable (r.v) X is said to follow Laplace distribution if its probability density function (pdf) is given by:

$$f(x) \ = \ \frac{1}{2 \theta} \ {e^-}^{\frac{|x- \mu|}{\theta}} \ \ \ \ \ ; \ - \infty \ < \ x \ < \ \infty, \ - \infty \ < \ \mu \ < \ \infty, \ \theta \ > \ 0$$

If X follows Laplace Distribution with parametersμ andθ, it is denoated as X ~ L (μ,θ).

The continious random variable representing longest life time of machines or objects follows Laplace distribution.

Ifμ = 0 andθ = 1, then the random variable X has a standard Laplace distribution defined as follows:

Definition :A random variable X is said to follow a standard Laplace distribution if its probability density function is

$$f(x) \ = \ \frac{1}{2} \ {e^-}^{|x|} \ \ \ \ \ ; \ - \ \infty \ < \ x \ < \ \infty.$$

If X has a standard Laplace distribution it is denoted as X ~ L (0,1).

Bibliography

Sukubhattu N.P. (2013). Probability & Inference - II. Asmita Books Publishers & Distributors (P) Ltd., Kathmandu.

Larson H.J. Introduction to Probability Theory and Statistical Inference. WileyInternational, New York.